OPEN CONTESTS PAGE
If you're willing to risk $50 or more on a problem, write me.
You don't have to pay *me* anything - just pay whoever solves your
problem!

Prize Problem -- $500. Claim: For an integer n
> 2, there are no solutions in the Gaussian integers to the equation xn+yn=zn.
I hereby offer $500 for a counterexample, with requirements |x y z|
> 0 and {x,y,z}
Gaussian Integers. (I have no interest in impossibility proofs).
Dave Rusin pointed me to a math-atlas
note.

Prize Puzzle. A $1.5 million prize for the first program
that can play Go at a professional level. This is the the Ing Chang-Ki Prize.

Prize Puzzle. A $1.0 million prize to the first one to
prove or disprove the Riemann Hypothesis. The Clay Mathematics Institute of Cambridge,
Massachusetts (CMI) has named seven “Millennium Prize Problems.”
The Board of Directors of CMI have designated a $7 million prize fund
for the solution to these problems, with $1 million allocated to
each. You can see more details at www.claymath.org. The full
list of problems: P versus NP, The Hodge Conjecture, The
Poincaré Conjecture, The Reimann Hypothesis, Yang-Mills Existence
and Mass Gap, Navier-Stokes Existence and Smoothness, and The Birch
and Swinnerton-Dyer Conjecture. On the Riemann Hypothesis, the
first hundred billion zeroes have been calculated by Zetagrid, without
any counterexamples so far.

Prize puzzle. A $100,000 prize
for the first verified prime number with more than 10,000,000
digits. See eff.org
for more details. www.mersenne.org
is the main solving organization. 233219313 - 1 would
be the first number to check. The smallest prime factor of
10^10000000! + 1 would also work.

Prize puzzle. A $100,000 prize for the first to solve The
Beal Conjecture. If Ax +By = Cz ,
where A, B, C, x, y and z are positive integers and x, y and z are all
greater than 2, then A, B and C must have a common factor.

Prize puzzle. ZetaGrid is offering the following four
prizes (see http://www.zetagrid.net/zeta/prizes.html
for more rules):
1. $10 (USD) will be awarded to the first person who discovers the
first two zeros which have a distance less than 10^-6, using the
software provided by ZetaGrid.
2. $100 (USD) will be awarded to the first person who discovers the
first two zeros which have a distance less than 10^-7, using the
software provided by ZetaGrid.
3. $1,000 (USD) will be awarded to the first person who discovers a
nontrivial zero which is not on the critical line using the software
provided by ZetaGrid and before the Riemann Hypothesis disproved.
4. Up to $1,000,000 (USD) award to the first 100 top producers of
ZetaGrid if Sebastian Wedeniwski wins the $1 million prize for a proof
of the Riemann Hypothesis from Clay Mathematics Institute by using
the results or the statistical summaries of ZetaGrid.

Prize puzzle. A $100 prize is
offered by Martin Gardner's for the first to make a 3x3 magic square
where all entries are different perfect squares. Or prove that
this is impossible. The best known construction is below, by Lee
Sallows. One diagonal fails to add up to the magic constant of
147^2. Adam Dewbery (and many others) proved
that the central square of a solution must have several properties
(I've added to it). Interesting things like 472+792
= 232+892 = 132+912 = 352+752
= 652+652 pop up during analysis. Near-magic
squares are completely characterized (near-magic squares are ones with
all rows, all columns, and one of the two diagonals adding to the magic
constant). For any congruent number n (an integer
that is the area of some rational right triangle), find 3 points on
the elliptic curve y2 = x3-nx, such that each point
is the double of another point on the curve. Each point
corresponds to a 3-term arithmetic progression. If these three
progressions are distinct, then they generate a near-magic square. See
John Robertson's paper Magic Squares of Squares in the October
1996 Mathematics Magazine, pp. 289 ff. This problem is now
famous enough that the only proofs I will consider are those published
in refereed mathematical journal.

Prize Puzzle. $50 for a complete solution. The Logistic Map, x -> a x(1-x).
At seanet.com,
exact answers are given for cases a=2 and a=4. The first one -- after n
iterations of x -> 2
x(1-x), the value of x will be (1 - Exp[2^n Log[1 - 2 x]])/2 . As a
tough problem (I don't have an answer), what is the exact formula for x -> 3 x(1-x)
after n iterations? If you can solve this, or if you know of existing
literature on related exact formulas, please write me. The below piece of
code shows the standard diagram, with the first hundred iterations
tossed out. You can see iteractive versions of the logistic map at Wolfram.com,Geometry
Center, Gallery
of Mathematics, University
of Tokyo, Burning
Chaos Page, Chaos
and Order, and Mathpost.
It's related to the Lyapunov Exponent. Nice explanations of this are
given by J C
Sprott, and the Chaos
Hypertextbook.

Prize Puzzle. How many pieces are necessary to make the
97 two tile combinations of Chaos Tiles
simultaneously? $50 to the first person that can do it with 50
Veks and 40 Kays, or fewer. (No-one entered the original contest,
which expired 7 Dec 2000)

There is no better prize for solving a mathematical puzzle than a puzzle itself. Create unique one of a kind puzzles with your favorite algorithm, veks or design.

Prize Puzzle. A $50 prize will go to the first to solve or
prove impossible Octiamond Solitaire. Solved by
Wei-Hwa Huang
At every stage of the game, each piece must be completely
supported. Use the full set of 66 octiamonds, without reusing
any.
Start: Put one octiamond piece down. This is level one.
Stage 2: Add an octiamond to level one. Put another
octiamond on top, starting level two.
Stage 3: Add an octiamond to level one, then add another to level
two, then put another on top to make level three.
...
Stage 11: Add an octiamond to level one, then add another to
level two, then add another on level three, ..., then put the last
octiamond on top to make level eleven.
"The algorithm used to solve it worked from the top of the "pyramid"
and searched for possible moves downwards. Retrograde analysis,
in a sense. Note that the longer pieces are at the top, while
the compact, squat pieces are at the bottom."

Petersen Graph Zome Challenge -- $100. My Petersen
Graph Zome challenge has been solved by David MacMahon. It's
not possible. If you have a plug-in like Cosmo, you can examine his VRML
constructions. His concept is described at his Analytic page. He put the
solving technique on his Petersen page.
The Problem: Arrange 10 nodes and 15 blue struts of the same length to
make the Petersen Graph (above). No struts may touch. Here
is a 14 strut construct by Wei-Hwa Huang. I've also found
several 14 strut constructions. A 15 strut construction may not
be possible. A computer proof of impossibility is also
acceptable.